Actual Questiona sample of iron ore weighing 2.8 g is to be analyzed for the percentage of iron. all of the iron in the sample is converted into iron oxide (Fe2O3) which has a mass of 1.0 g. what is the percentage of iron in the ore?
well I'm a sophomore in high school and i kinda taught my self this(requirement for ap chem) so I'll show u my process, but im less than 50% that i used the right process. 55.85g Fe/ 1mol Fe * 2molFe/ 1 mol Fe2O3 * 1mol Fe2O3/159.79g Fe2O3 * 100 69.94 % Fe
The atomic weight of Fe 55.85 g/mol, so 2 moles would weigh 111.70 g The atomic weight of O 16 g/mol, so 3 moles would weigh 48 g This makes a total of 159.70g, but you are told you have 1.0 g therefore the proportion of this which is Fe 111.70 / 159.70 * 1.0 g 0.699 g or 0.7 g to the same accuracy 0.7 g as a percentage of the 2.8 g sample of iron ore 0.7 / 2.8 * 100 25.0%
well I'm a sophomore in high school and i kinda taught my self this(requirement for ap chem) so I'll show u my process, but im less than 50% that i used the right process. 55.85g Fe/ 1mol Fe * 2molFe/ 1 mol Fe2O3 * 1mol Fe2O3/159.79g Fe2O3 * 100 69.94 % Fe
The atomic weight of Fe 55.85 g/mol, so 2 moles would weigh 111.70 g The atomic weight of O 16 g/mol, so 3 moles would weigh 48 g This makes a total of 159.70g, but you are told you have 1.0 g therefore the proportion of this which is Fe 111.70 / 159.70 * 1.0 g 0.699 g or 0.7 g to the same accuracy 0.7 g as a percentage of the 2.8 g sample of iron ore 0.7 / 2.8 * 100 25.0%
